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$\mathbb{C f} a^x=b^y=c^z=d^w$, the value $\alpha, x\left(\frac{1}{y}+\frac{1}{z}+\frac{1}{w}\right)$ is
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The correct answer is:
$\log _e(k c d)$
Given, $a^x=b^y=c^z=d^w$
$\begin{aligned} & \Rightarrow x=y \log _a b=z \log _a c=w \log _a d \\ & \Rightarrow y=\frac{x}{\log _a b}, z=\frac{x}{\log _a c}, w=\frac{x}{\log _a d}\end{aligned}$
Now, $\begin{aligned} & x\left(\frac{1}{y}+\frac{1}{z}+\frac{1}{w}\right) \\ = & x\left[\frac{\log _a b}{x}+\frac{\log _a c}{x}+\frac{\log _a d}{x}\right] \\ = & \frac{x}{x}\left[\log _a b c d\right] \\ = & \log _a(b c d)\end{aligned}$
$\begin{aligned} & \Rightarrow x=y \log _a b=z \log _a c=w \log _a d \\ & \Rightarrow y=\frac{x}{\log _a b}, z=\frac{x}{\log _a c}, w=\frac{x}{\log _a d}\end{aligned}$
Now, $\begin{aligned} & x\left(\frac{1}{y}+\frac{1}{z}+\frac{1}{w}\right) \\ = & x\left[\frac{\log _a b}{x}+\frac{\log _a c}{x}+\frac{\log _a d}{x}\right] \\ = & \frac{x}{x}\left[\log _a b c d\right] \\ = & \log _a(b c d)\end{aligned}$
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